In Fitzpatrick's Newtonian Dynamics book on the Coriolis force, he states

\begin{align} v_{x'}&\simeq V_0\cos\theta-2\Omega t V_0\sin\lambda~\sin\theta \tag{433}\\ v_{y'}&\simeq-V_0\sin\theta-2\Omega t V_0\sin\lambda~\cos\theta \tag{434} \end{align} To lowest order in $\Omega$, the above equations are equivalent to \begin{align} v_{x'}&\simeq V_0\cos\left(\theta+2\Omega t\sin\lambda \right)\tag{435}\\ v_{y'}&\simeq -V_0\sin\left(\theta+2\Omega t\sin\lambda \right)\tag{436} \end{align}

I simply don't get how the above equations are equivalent. In fact, if I take a magnitude (of the total vector) they are different. I also don't understand what "To lowest order in $\Omega$" means here.


Note that Fitzpatrick states towards the beginning,

The following solution method exploits the fact that the Coriolis force is much smaller in magnitude that the force of gravity: hence, $\Omega$ can be treated as a small parameter

Generally, when statements like that are made, powers (greater than 1) of the term in question are considered to be zero: $$ \Omega\ll1; \quad \Omega^n\,\approx0\,\,\forall\,\,n>1 $$ Thus, $$ a\Omega+b\Omega^2\approx a\Omega $$ for small $\Omega$.

In the small angle approximation, we have \begin{align} \sin\theta&\approx\theta\\ \cos\theta&\approx1-\frac{\theta^2}2\\ \tan\theta&\approx\theta \end{align} We can apply this to your first $v_{x'}$ equation: \begin{align} v_{x'}&\approx V_0\left(\cos\theta-2\Omega\sin\lambda \sin\theta t\right)\\ &\approx V_0\left(1-\frac{\theta^2}2-2\Omega\sin\lambda\theta t\right) \end{align} You can then complete the square to finish the solution, ignore the $\Omega^2$ term and end up with Equation (435) that the author gives.

  • $\begingroup$ I've completed the square and get $Vo((1-\Omega\theta sin\lambda t )^2 -\frac{\theta^2}{2})$ I cant come up with the final equation, hehe. $\endgroup$ – DLV Nov 27 '14 at 4:10
  • $\begingroup$ You should only be considering the terms with respect to $\theta$, so complete the square with $\theta^2/2+2\Omega\sin\lambda\theta t$. $\endgroup$ – Kyle Kanos Nov 27 '14 at 4:16
  • $\begingroup$ Wow. I'm just impressed. Is this a famous derivation or something? When does this stop being valid? How much time has to pass? Thanks. $\endgroup$ – DLV Nov 27 '14 at 4:30
  • 1
    $\begingroup$ Famous? Probably not, I cheated and expanded (435) and recognized the difference between the two relations as a completing the square :). It stops being a valid relation when the second term is no longer small (either $\Omega\approx1$ or $t$ being some nominal value). $\endgroup$ – Kyle Kanos Nov 27 '14 at 4:33

Ahh, Richard Fitzpatrick. Great guy.
Ok, If you start with the second set of expressions, use the appropriate double-angle-formula and then assume the "angle" $2\Omega \sin\lambda t$ is small (note that the $t$ is not within the sin function!), you get the first expressions, e.g.

$$\cos(\theta+\phi) = \cos\theta\cos\phi - \sin\theta\sin\phi,$$ and then for small $\phi$, use the first-order terms in the Taylor expansion for the trig functions, i.e. $\cos\phi \simeq 1$, and $\sin\phi\simeq \phi$. In your case, $\phi = 2\Omega\sin\lambda t$, where again the $t$ is intended as a (linear) multiplicative factor and not inside the $\sin$ function.

Similarly, $$\sin(\theta+\phi) = \sin\theta\cos\phi + \cos\theta\sin\phi,$$ and you can take if from there...;-)

Woops, looks like Kyle beat me to the punch while I was typing this in!

  • $\begingroup$ Out of curiosity, could Fitzpatrick's books become non-free any time? They are too good. I guess I'll download them for fear of this. Hehe. $\endgroup$ – DLV Nov 27 '14 at 6:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.